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2^(4-1)^2

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by hwiya320 » Sat Aug 01, 2009 12:59 pm
do what's in the parenthesis first, then you have 2^3^2 over 2^1

or 2^ (3^2) which equals to 2^9 / 2^1 which equals to 2^8.

you're multiplying 3 with 2, rather than squaring the 3.
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by alexdallas » Sat Aug 01, 2009 1:05 pm
got it.

completely missed the parantheses.
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by polarbear222 » Sat Aug 01, 2009 2:25 pm
this one tripped me up to. i still don't really understand what the difference is between:

4^(3)^2 and (4^3)^2

so in the first its 4^9 and second its 4^6? i'm not sure why this doesn't make sense to me. what is the order of operations rule i am missing?
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by alexdallas » Sat Aug 01, 2009 2:27 pm
its PEMBAS.

parantheses- always come first.
so if you do the parantheses, that's (4-1)squared, which is 3squared, which is 9.
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by nk_81 » Sun Mar 06, 2011 1:48 am
I seek clarification....

as per the attachment 2^(4-1)^2 is different from (2^(4-1))^2 ...right? The answer of the latter is going to be 2^5...is it?
NK
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by Anurag@Gurome » Sun Mar 06, 2011 3:30 am
nk_81 wrote:I seek clarification....

as per the attachment 2^(4-1)^2 is different from (2^(4-1))^2 ...right? The answer of the latter is going to be 2^5...is it?
Yes, you're correct.

2^[(4 - 1)^2] = 2^9 ...... BUT ...... [2^(4 - 1)]^2 = 2^6
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by nk_81 » Sun Mar 06, 2011 5:36 am
Anurag@Gurome wrote:
nk_81 wrote:I seek clarification....

as per the attachment 2^(4-1)^2 is different from (2^(4-1))^2 ...right? The answer of the latter is going to be 2^5...is it?
Yes, you're correct.

2^[(4 - 1)^2] = 2^9 ...... BUT ...... [2^(4 - 1)]^2 = 2^6
In this case, we just need to watch out for the parathesis then. The position of the parathesis will decide the solution for such questions. Right?
NK
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by RyanP » Tue Apr 03, 2012 7:10 pm
Why do you just automatically take out the parenthesis after doing 4-1. Shouldn't it still be 2^(3)^2?
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by RyanP » Thu Apr 05, 2012 6:11 pm
RyanP wrote:Why do you just automatically take out the parenthesis after doing 4-1. Shouldn't it still be 2^(3)^2?
Anybody?
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by Bill@VeritasPrep » Thu Apr 05, 2012 6:55 pm
RyanP wrote:Why do you just automatically take out the parenthesis after doing 4-1. Shouldn't it still be 2^(3)^2?
Yes, it should be.
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by RyanP » Fri Apr 06, 2012 11:34 am
Bill@VeritasPrep wrote:
RyanP wrote:Why do you just automatically take out the parenthesis after doing 4-1. Shouldn't it still be 2^(3)^2?
Yes, it should be.
If it should be that, then why is the answer 2^8 rather than 2^5? OG has it as 2^8
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by seal4913 » Fri Apr 06, 2012 11:46 am
RyanP wrote:
Bill@VeritasPrep wrote:
RyanP wrote:Why do you just automatically take out the parenthesis after doing 4-1. Shouldn't it still be 2^(3)^2?
Yes, it should be.
If it should be that, then why is the answer 2^8 rather than 2^5? OG has it as 2^8
2^(3)^2, the parenthesis is 3^2 which is 9 and gives (2^9)/(2^1) which is still 2^8
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by Bill@VeritasPrep » Fri Apr 06, 2012 12:10 pm
^Exactly. The parentheses tell us which base to raise to its exponent first.
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by KateG1302 » Thu Oct 25, 2012 4:10 pm
Hi all,

sorry to dig this thread out again. I am not 100% clear on how the rules are with regards to exponents of exponents - I thought it was always just a question of multiplying them (ie 2^4^3 = 2^12) but now it seems that the rules have changed.

I don't quite understand where the brackets have to be to do the 3^2 in this case instead of 3*2. It would be great if someone could explain WHEN to use ^ and when *.

I kind of got from the previous explanation that when the exponent and its exponent (in this case: (4-1)^2) are in brackets, then I need to do ^ - i.e. 3^2. And if the base is in a bracket with the "first exponent" (in this case: 2^(4-1)) then I need to multiply the first and the second exponent (ie (4-1)^2 = 3*2).

Is this always the rule? Can someone confirm if I'm right here please?

Thanks in advance!

Kate
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